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Equation of a Sphere, Plus Center and Radius

by Krista King

Sphere's are the 3D representations of circles. The equation of a sphere is similar to that of a circle, but with an extra variable for the extra dimension. $$(x-h)^2+(y-k)^2+\color{Red}{(z-l)^2}=r^2$$ In this equation, $r=\text{radius}$. The coordinate $(h,k,l)$ tells us where the center of the sphere is.

In this video, Krista goes over two types of problems you might encounter when dealing with equations of spheres.

Find Equation Given Point and Center

Suppose you have a sphere whose center is $C(3,8,1)$ and it passes through the point $(4,3,-1)$. What is its equation?

First, we are given the center of the sphere: $(3,8,1)$. Let's plug this point into our sphere equation. $$\begin{align} (x-h)^2+(y-k)^2+(z-l)^2&=r^2\\ (x-\color{Red}{3})^2+(y-\color{Red}{8})^2+(z-\color{Red}{1})^2&=r^2 \end{align}$$

We're actually almost done! We also know the sphere passes through the point $(4,3,-1)$. We can plug this in for $(x,y,z)$ and solve for $r^2$. $$\begin{align} (x-3)^2+(y-8)^2+(z-1)^2&=r^2\\ (\color{Blue}{4}-3)^2+(\color{Blue}{3}-8)^2+(\color{Blue}{-1}-1)^2&=r^2\\ (1)^2+(-5)^2+(-2)^2&=r^2\\ 1+25+4&=r^2\\ 30&=r^2 \end{align}$$

Now that we know $r^2$ and the center, we know the equation of the sphere. $$\boxed{(x-3)^2+(y-8)^2+(z-1)^2=30}$$

Find Center and Radius Given an Equation

Suppose you have a sphere with the equation $x^2+y^2+z^2-2x-4y+8z=15$. What is the radius and center of this sphere? To solve these problems, we want to start by grouping our variables together. $$\begin{align} x^2+y^2+z^2-2x-4y+8z&=15\\ (x^2-2x)+(y^2-4y)+(z^2+8z)&=15 \end{align}$$

Next, we want to complete the square for each grouping. When completing the square, we want to take the coefficient in front of the variable and halve it, square it, and add it on to both ends. For example, in our equation above, to complete the square for the $x$ variable, we take the $-2$ and halve it, $\frac{-2}{2}=-1$, square it, $(-1)^2=1$, and add it to both sides. $$\begin{align} (x^2-2x)+(y^2-4y)+(z^2+8z)&=15\\ \left(x^2-2x+\color{Green}{\left(\frac{-2}{2}\right)^2}\right)+(y^2-4y)+(z^2+8z)&=15+\color{Green}{\left(\frac{-2}{2}\right)^2}\\ (x^2-2x+1)+(y^2-4y)+(z^2+8z)&=15+1\\ (x^2-2x+1)+\left(y^2-4y+\color{Green}{\left(\frac{-4}{2}\right)^2}\right)+(z^2+8z)&=15+1+\color{Green}{\left(\frac{-4}{2}\right)^2}\\ (x^2-2x+1)+(y^2-4y+4)+(z^2+8z)&=15+1+4\\ (x^2-2x+1)+(y^2-4y+4)+\left(z^2+8z+\color{Green}{\left(\frac{8}{2}\right)^2}\right)&=15+1+4+\color{Green}{\left(\frac{8}{2}\right)^2}\\ (x^2-2x+1)+(y^2-4y+4)+(z^2+8z+16)&=15+1+4+16\\ (x^2-2x+1)+(y^2-4y+4)+(z^2+8z+16)&=36\\ \end{align}$$

Once we've completed the square, we can factor each grouping. $$\begin{align} (x^2-2x+1)+(y^2-4y+4)+(z^2+8z+16)&=36\\ (x-1)(x-1)+(y-2)(y-2)+(z+4)(z+4)&=36\\ (x-1)^2+(y-2)^2+(z+4)^2&=36 \end{align}$$

We now have our equation in a form we can understand. The center of the sphere are number being subtracted. $$(x-\color{Red}{1})^2+(y-\color{Red}{2})^2+(z-\color{Red}{(-4)})^2=36$$ The center is at $(1,2,-4)$.

The radius is the square root of the number at the end. $$r=\sqrt{36}=6$$ The sphere has a radius of 6.

Let's think about a sphere for a bit. For a concrete example, we could consider, say, a beach ball or a marble or a tennis ball or any such spherical object of your liking.

If your preferred ball-shaped item is perfectly round, then every single point on its surface would be equally distanced from its center. This distance can be referred to as the radius (just like in a circle!). Moreover, since this radius is always the same for all points on the sphere, we can create an equation to mathematically represent this sphere.

If we think back to working with circles, remember that a circle with center $(a,b)$ and radius $r$ is represented by the equation $$(x-a)^2 + (y-b)^2 = r^2.$$ A circle lives in two dimensions and a sphere in three. The two shapes both represent the set of all points equidistant from a single center, so the equation can easily be generalized to three dimensions!

But, what is this equation?

It turns out that the equation of a sphere with center $(a,b,c)$ and radius $r$ has the equation $$(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2.$$ This formula has its origins in the general distance formula, which lets you calculate the distance between any two specific points, even in 3D space.

For some practice, consider the sphere denoted by the equation $$(x-4)^2 + (y-1)^2 + (z-9)^2 = 16.$$

What if the equations we get are a little more complicated? What if the $(x-a)^2$, $(y-b)^2$, and $(z-c)^2$ were expanded and shuffled around?

The video below by patrickJMT shows how to find our crucial pieces of information — the center and radius — by completing the square. (Click here to open in a new tab.)

With our new tools in hand, let's move on to a final bit of practice. Consider the sphere denoted by the equation $$x^2 + y^2 + z^2 = 2x + 18y + 2z - 79.$$ Manipulate the equation into a more useful form, then answer the following questions.

Good job! If you're intrigued by spheres and want to challenge yourself, there's an entire type of geometry built around it: spherical geometry.

Sphere Intersection

So we know how to graph a sphere. But how can we tell if two spheres intersect?

Let's start out with some smaller dimensions.

Note: Not everyone agrees on the definition of a sphere, but throughout this explanation, a sphere will include everything inside. (Sometimes that would be called a ball instead.) When we write an equation like $x^2+y^2+z^2 = 1$, that will refer to the points on the surface of the sphere; the inequality $x^2+y^2+z^2 \leq 1$ refers to everything within or on the sphere.

Zero-Dimensional Spheres???

We've gone too small! But let's try reasoning out what happens anyway for the sake of being good mathematicians.

A "zero-dimensional sphere" is really just a point; we know two points "intersect" when they are exactly the same point!

One-Dimensional Spheres

A "one-dimensional sphere" is a line (or, to be more specific, a line segment); two segments will intersect at at most one point, unless they are part of the same extended line (like two intervals on a number line). You can read more about intersecting lines here.

One cool way to think about lines as one-dimensional spheres is to consider equations such as $(x-a)^2=r^2$ for a "sphere" centered at a point $a$ with radius $r$ on a line. Then we can visualize a sphere on a line like so:

Then how do we check if two if these segments intersect? Well, we can look at the center points and radii of two such segments.

Suppose we have two segments: segment $A$ with radius $r$ centered at point $a$ and segment $B$ with radius $q$ centered at point $b$. If any part of a segment $A$ overlaps with segment $B$, then they intersect. Symbolically, if the intervals $[a-r, a+r]$ and $[b-q, b+q]$ overlap, then we know our one-dimensional spheres intersect.

Sphere $A$ and Sphere $C$ intersect because they represent the line segments $[-3, 3]$ and $[-1, 9]$, so they overlap in $[-1, 3]$. Sphere $B$ is the line segment $[10, 20]$, so it doesn't overlap with either $A$ or $C$.

With that done, we're ready to move up a dimension!

Two-Dimensional Spheres (A.K.A. Circles)

Circle time! We define a circle with radius $r$ at a point $(a,b)$ with the equation $(x-a)^2+(y-b)^2=r^2$. So a two-dimensional sphere looks like this:

So how do we check if two of these circles intersect? Well, we can look at the center points and radii of two such circles.

Suppose we have two circles: circle $A$ with radius $r$ centered at point $(a,b)$ and circle $B$ with radius $q$ centered at point $(c, d)$. If any part of $A$ overlaps with $B$, then they intersect.

Symbolically, if the equations (or rather, inequalities) $$(x-a)^2+(y-b)^2\leq r^2$$ $$(x-c)^2+(y-d)^2\leq q^2$$ have solutions, then we know our two-dimensional spheres intersect.

More simply, we want to see if there are any points radius distance away from the center of either circle.

Sphere $B$ and $C$ share the point $(0, 4)$! Moving on...

Three-Dimensional Spheres

Finally, we've made it to the question we actually want to answer! We know we can represent a sphere with radius $r$ centered at the point $(a, b, c)$ has the equation $(x-a)^2+(y-b)^2+(z-c)^2=r^2$.

Let's try building up our intuition with our smaller examples. We want two intersecting spheres to share points that are their respective radius distance apart from their centers. Try it out for yourself!

Now you may think we're done, but we can go higher!

Four-Dimensional Spheres???

Here's one of the best parts of math: we can generalize the patterns we find in small examples to work with spaces we might not even be able to concretely visualize!

If the equation for one-dimensional spheres is $(x-a)^2=r^2$, for two-dimensional spheres is $(x-a)^2+(y-b)^2=r^2$, and for three-dimensional spheres is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$, then it follows that an equation for a four-dimensional sphere - whatever that is - must be: $$(x-a)^2+(y-b)^2+(z-c)^2+(w-d)^2=r^2$$

A sphere is the set of all points a given distance (its "radius") from a given point (the "center" of the sphere). As the sphere is a generalization of the circle, it stands to reason their equations would look similar.

Indeed, recall that for a circle with center $(a,b)$ and radius $r$, the equation is:

$(x-a)^2+(y-b)^2=r^2$

We can derive a similar result for three dimensions using the distance formula. The sphere of radius $r$ centered at $(a,b,c)$ is simply the set of all points $(x,y,z)$ such that the distance between $(x,y,z)$ and $(a,b,c)$ is $r$. By the aforementioned distance formula, this is:

$\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}=r$

so the equation for the sphere is just:

$\boxed{(x-a)^2+(y-b)^2+(z-c)^2=r^2}$

Note the similarity with the circle's equation! Can you figure out how this would generalize to higher dimensional "spheres"?